The maximum value of \(\mathrm{Z}=3 x+5 \mathrm{y}\), subject to \(3 x+2 \mathrm{y} \leq 18, x \leq 4, \mathrm{y} \leq 6\)
\(x, y \geq 0\) is

Point of intersection of \(x=4\) and \(3 x+2 y=18\) is \(Q \equiv(4,3)\) Point of intersection of \(\mathrm{y}=6\) and \(3 \mathrm{x}+2 \mathrm{y}=18\) is \(\mathrm{P} \equiv(2,6)\) Point \(\mathrm{D} \equiv(4,0)\) and \(\mathrm{C} \equiv(0,6)\) are as shown. The feasible region of th given L.P.P. is shaded portion CPQ D O. We have to maximize \(Z=3 x+5 y\) Now, \(\mathrm{Z}\) at \(\mathrm{C}(0,6)=3(0)+5(6)=30\)
\(\mathrm{Z}\) at \(\mathrm{P}(2,6) \quad=3(2)+5(6)=36\)
\(\mathrm{Z}\) at \(\mathrm{Q}(4,3)=3(4)+5(3)=27\)
\(Z\) at \(D(4,0)=3(4)+5(0)=12\)
\(\mathrm{Z}\) at \(\mathrm{O}(0,0)=3(0)+5(0)=0\)
Clearly the maximum value of \(\mathrm{Z}\) is 36 at \(\mathrm{P}(2,6)\)